Let $ A' $ be the reflection of $ A $ in $ BC $ and $ I, $ $ I_A $ be the incenter, A-excenter of $ \triangle ABC, $ respectively. Since $ \odot (II_BI_C) $ is the image of $ BC $ under the inversion with center $ I $ that swaps $ (A,I_A), $ $ (B,I_B), $ $ (C,I_C), $ so $ IA' $ passes through the image $ P^* $ of $ I_A $ under the inversion WRT $ \odot (II_BI_C). $ Similarly, we can prove $ P^* $ lies on $ I_CB', $ $ I_BC', $ so $ P^* $ $ \equiv $ $ P $ $ \Longrightarrow $ $ P $ $ \in $ $ IA'. $ Notice that the circumcircle $ \odot (O) $ is the 9-point circle of $ \triangle II_BI_C, $ so the perpendicular bisector $ \tau $ of $ PI_A $ is the polar of $ I_A $ WRT $ \odot (O). $ Let $ T $ be the midpoint of $ PI_A $ and $ J $ be the intersection of $ \tau, $ $ BC. $ Since $ J $ lies on the perpendicular bisector of $ AI_A, $ so $ A, $ $ A', $ $ I_A, $ $ P, $ $ P_A $ lie on a circle with center $ J, $ hence notice $ \triangle OAT $ $ \stackrel{-}{\sim} $ $ \triangle OI_AA $ we get \begin{align*} \measuredangle (AT,AI) &= \measuredangle (AT,AO) + \measuredangle (AO,AI) \\ &= \measuredangle (OI_A, AI) + \measuredangle (AI,\perp BC) \\ &= \measuredangle (PI_A, PP_A) \\ &= \measuredangle (AI,AP_A) \end{align*}$ \Longrightarrow $ $ AP_A $ passes through the isogonal conjugate $ \widetilde{T} $ of $ T $ WRT $ \triangle ABC. $ Analogously, we can prove $ \widetilde{T} $ lies on $ BP_B, $ $ CP_C. $ ____________________________________________________________ Remark : Generally, if $ D $ is the second intersection of $ AP $ with $ \odot (BPC) $ and $ X $ is the reflection of $ D $ in $ BC, $ then $ AX $ passes through the antigonal conjugate of $ P $ WRT $ \triangle ABC. $

Attachments:

I see a generalization as following Let $ABC$ be a triangle and $P,Q$ are two points on Neuberg cubic. $PA,PB,PC$ cuts the circles $(PBC),(PCA),(PAB)$ again at $P_a,P_b,P_c$. $QA,QB,QC$ cuts the circles $(QBC),(QCA),(QAB)$ again at $Q_a,Q_b,Q_c$. Then $P_aQ_a,P_bQ_b,P_cQ_c$ are concurrent at a point on this Neuberg cubic.

buratinogigle wrote: I see a generalization as following Let $ABC$ be a triangle and $P,Q$ are two points on Neuberg cubic. $PA,PB,PC$ cuts the circles $(PBC),(PCA),(PAB)$ again at $P_a,P_b,P_c$. $QA,QB,QC$ cuts the circles $(QBC),(QCA),(QAB)$ again at $Q_a,Q_b,Q_c$. Then $P_aQ_a,P_bQ_b,P_cQ_c$ are concurrent at a point on this Neuberg cubic. Generalization : Given a $ \triangle ABC $ and a point $ P $ at infinity. Let $ \Omega $ be the pivotal isogonal cubic of $ \triangle ABC $ with pivot $ P $ and let $ U, $ $ V $ be the points lying on $ \Omega. $ Let $ U^*, $ $ V^* $ be the isogonal conjugate of $ U, $ $ V $ WRT $ \triangle ABC, $ resp. and let $ \triangle U_aU_bU_c, $ $ \triangle V_aV_bV_c $ be the circlecevian triangle of $ U, $ $ V $ WRT $ \triangle ABC, $ respectively. Then $ U^*V^*, $ $ U_aV_a, $ $ U_bV_b, $ $ U_cV_c, $ $ \Omega $ are concurrent. Proof : Let $ P^* $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ and let $ W $ be the third intersection of $ UV $ with $ \Omega. $ From Property of Cevian quotient (post #3, #4) $ \Longrightarrow $ $ AP^*, $ $ BP^*, $ $ CP^*, $ $ PP^* $ tangents to $ \Omega $ at $ A, $ $ B, $ $ C, $ $ P, $ resp. and $ P^*, $ $ W, $ $ (P/W) $ are collinear. Furthermore, $ (P/W) $ lies on $ \Omega $ (Isogonal conjugate and Cevian quotient (post #4)), so by Cayley-Bacharach theorem for $$ \left( \Omega \ , \ \overline{PUU^*} \cup \overline{PVV^*} \cup \overline{P^*W(P/W)} \ , \ \overline{PPP^*} \cup \overline{UVW} \cup \overline{U^*V^*} \right) $$we get $ (P/W) $ lies on $ U^*V^*. $ Notice $ U_a, $ $ V_a, $ $ U_b, $ $ V_b, $ $ U_c, $ $ V_c $ lie on $ \Omega, $ so by Cayley-Bacharach theorem for \begin{align*} \left( \Omega \ , \ \overline{AUU_a} \cup \overline{AVV_a} \cup \overline{P^*W(P/W)} \ , \ \overline{AAP^*} \cup \overline{UVW} \cup \overline{U_aV_a} \right) \\ \left( \Omega \ , \ \overline{BUU_b} \cup \overline{BVV_b} \cup \overline{P^*W(P/W)} \ , \ \overline{BBP^*} \cup \overline{UVW} \cup \overline{U_bV_b} \right) \\ \left( \Omega \ , \ \overline{CUU_c} \cup \overline{CVV_c} \cup \overline{P^*W(P/W)} \ , \ \overline{CCP^*} \cup \overline{UVW} \cup \overline{U_cV_c} \right) \end{align*}we get $ (P/W) $ lies on $ U_aV_a, $ $ U_bV_b, $ $ U_cV_c, $ hence we conclude that $ U^*V^*, $ $ U_aV_a, $ $ U_bV_b, $ $ U_cV_c, $ $ \Omega $ are concurrent.

nice problem my solution: Let $A'$ be the reflection of $A$ in $BC$ and let $I,I_A$ be the incenter, A-excenter of $ABC$. Let $P'$ be the image of $I_A$ under inversion about $(II_BI_C)$. As in TelvCohl's solution, we want to prove $P'\in IA', I_CB', I_BC'$. First we prove $P\in IA'$. Let $O, I', O'$ be the circumcenter of $ABC$, the circumcenter of $I_AI_BI_C$, and the circumcenter of $II_BI_C$. Then clearly $I,O,I'$ are collinear, hence $OI$ is orthogonal to $(I_AI_BI_C)$. Applying $\sqrt{bc}$-inversion tells us that $(AA'I_A), (II_BI_C)$ are orthogonal so we know $P' \in (AA'I_A), O_1I_A$. But now we have $\angle O_1IP' =\angle O_1I_AI = \angle PA'A = 180^{\circ} - \angle A'IO_1$ because $IO_1\perp BC$ and $O_1I, (P'II_A)$ are tangent, implying that $P',I,A'$ are collinear as desired. Next we prove $I_B,P', C'$ are collinear. Angle-chasing yields $I_BO_1\perp AB$, hence $\angle O_1I_BC' = \angle I_BC'C$. By $\sqrt{ab}$-inversion, this equals $\angle CI_AO = \angle I_BI_AP' = \angle O_1I_BP'$, so $I_B,P',C'$ are collinear as desired. Similarly, $I_C,P', B'$ are collinear, so $P=P'$. Now let $B_2$ be the reflection of $I_C$ over $AC$ and define $C_2$ similarly. It's clear from the reflexive definition of $P_B,P_C$ that $P_B\in BB_2, P_C\in CC_2$ so it suffices to show $X=BB_2\cap CC_2$ lies on $AP_A$. Clearly $\angle C_2AB = \angle CAB_2 = 90^{\circ} + 0.5A$, and $\frac{C_2A}{AB} = \frac{I_BA}{AB} = \frac{AC}{I_CA} = \frac{AC}{AB_2}$, hence $\triangle C_2AB\sim \triangle CAB_2$, so by the spiral similarity center lemma we deduce $A \in (XBC_2), (CXB_2)$. Let $I_1$ be the reflection of $I$ over $BC$, so that $P,I,A'$ collinear implies $P_A, I_1, A$ are collinear, and it suffices to show $I_1$ lies on $AX$. We know $\angle AXB = 180^{\circ} - \angle AC_2B = 180^{\circ} - \angle AI_BB = 180^{\circ} - 0.5C$, and similarly $\angle AXC = 180^{\circ} - 0.5B$ so that $\angle BXC = 90^{\circ} - 0.5A$, implying that $X \in (BI_1C)$. Now clearly $\angle BXA = 180^{\circ} - 0.5C = 180^{\circ} - \angle BCI_1 = 180^{\circ} - \angle BXI_1$, so $A,X,I_1$ are collinear as desired. Note: can someone please elaborate on TelvCohl's proof that $P* \in I_CB', I_BC', IA'$? The proof I gave was the fastest one I could come up with

LEMMA:1 In a $\triangle ABC$ with circumcenter $O$ and orthocenter $H$ let $(P,Q)$ be isogonal conjugates wrt $\triangle ABC$ let $P_A,P_B,P_C$ be the reflection of $P$ in $BC,CA,AB$ if $AP_A,BP_B,CP_C$ are concurrent at $X$ let $X^*$ be the isogonal conjugate of $X$ wrt $\triangle ABC$ then $O,P,X^*$ are collinear and $P,Q,X$ are collinear. PROOF: Clearly $\{P,Q\}$ are Orthology centers of $\triangle ABC,\triangle P_AP_BP_C$ so by sondat's theorem we have $P,Q,X$ are collinear now let $H^*$ be the orthocenter of $\triangle P_AP_BP_C$ now we have $P_AH^*\parallel AQ,AH\parallel P_AP,AX\equiv P_AX$ so from Collinear with isogonal conjugate and circumcenter(post #2,LEMMA) we have $A,B,C,Q,H,X$ lie on a conic so taking isogonal conjugation wrt $\triangle ABC$ we have $P,O,X^*$ are collinear this proves the lemma. LEMMA:2 In a $\triangle ABC$ let $N,H$ be the nine point center and orthocenter and let $H^*$ be the inverse of $H$ wrt $\odot( ABC)$ then $N,H^*$ share the same orthotransversal wrt $\triangle ABC$. PROOF: Let $O$ be the circumcenter of $\triangle ABC$ and let $AH$ meet $BC,\odot (ABC)$ at $D,X$ so Now $HN.HH^*=\frac{\text{Pow}(H,\odot (ABC))}{2}=\frac{HA.HX}{2}=HA.HD$ so we have $A,D,N,H^*$ are cyclic so we conclude that $N,H^*$ have share same orthotransversal . LEMMA:3 In a $\triangle ABC$ let $N$ be the nine point center and $\triangle DEF$ be the orthic triangle then orthotransversal of $N$ wrt $\triangle ABC$ is perpendicular to euler line of $\triangle DEF$. PROOF: Let $H$ be the orthocenter of $\triangle DEF$ from Two triangles share the same orthocenter(post#2,LEMMA) we have $NH$ is perpendicular to orthotransversal of $N$ wrt $\triangle ABC$ so this proves the lemma$.$ ____________________________________________ Now coming back to the problem let $\mathcal{N}$ be the neuberg cubic of $\triangle ABC$ and let $\Omega,\mathcal{J}$ be the neuberg cubic, Jerabek hyperbola of $\triangle I_AI_BI_C$ where $I_A$ is the $A-$excenter and let $A'$ be the reflection of $A$ in $BC$ and let $I$ be the incenter of $\triangle ABC$ let $Be$ be the Bevan point of $\triangle ABC$ let $Be^*$ be the antigonal conjugate of $Be$ wrt $\triangle I_AI_BI_C$ and let $I^*$ be the inverse of $I$ wrt $\odot (I_AI_BI_C)$ . Now let $I_ABe\cap \odot (BeI_BI_C)=D$ and let $I_A^*,D^*$ be the reflection of $I_A,D$ in $I_BI_C$ similarly define $E^*,F^*$ now since the inversion centered at $I_A$ with power $\sqrt{I_AI_B.I_AI_C}$ followed by reflection in the bisector of $\angle I_BI_AI_C$ maps $I\mapsto D,Be\mapsto I_A^*$ so $IBe\parallel I_A^*D$ so $I_A^*D$ passes through $\infty_{IBe}$ and since $(I_AD^*,I_A^*D)$ are reflections in $I_BI_C$ and since $\infty_{IBe}$ lies on $\Omega$ we have that $I_AD^*,I_BE^*,I_CF^*$ are concurrent at say $L$ now from LEMMA:1 we have $L$ lies on $\mathcal{J}$ and also $L,\infty_{IBe},$ $\infty_{IBe}^*$ the isogonal conjugate of $\infty_{IBe}$ wrt $\triangle I_AI_BI_C$ which is clearly the fourth intersection of $\odot (I_AI_BI_C)$ and $\mathcal{J}$ now note that $L$ lies on $\mathcal{J}$ and $L\infty_{IBe}^*\parallel IBe$ so $\implies L\equiv Be^*$ .On the other hand $\triangle I_ABC\cup A'\sim \triangle I_AI_BI_C\cup D^*$ so we have $I_AA',I_BB',I_CC'$ are concurrent at the isogonal conjugate of $Be^*$ wrt $\triangle I_AI_BI_C$ since $Be$ is the antigonal conjugate of $Be$ wrt $\triangle I_AI_BI_C$ so we have $I_AA',I_BB',I_CC'$ are concurrent at $I^*$ . Now let $O$ be the circumcenter of $\triangle ABC$ from LEMMA:2 we know orthotransversal of $I^*,O$ wrt $\triangle I_AI_BI_C$ coincide so from the link mentioned in LEMMA:3 and from LEMMA:3 we have the line joining $I^*$ and $I'$ the isogonal conjugate of $I^*$ wrt $\triangle ABC$ is parallel to $OH$ so we have that $I^*$ lies on $\mathcal{N}$ now the involution mapping $I_B\mapsto I_C$ and $B'\mapsto C'$ is the $\sqrt{bc}$ inversion so $(P,I^*)$ are images under $\sqrt{bc}$ inversion if $Q$ is the image of $I'$ under $\sqrt{bc}$ inversion then since $P,Q$ are isogonal conjugates wrt $\triangle ABC$ and $I^*I'\parallel PQ$ we have that $P,Q$ also lie on $\mathcal{N}$ so $\implies AP_A,BP_B,CP_C$ are concurrent so we are done $\qquad \blacksquare$

Another short solution , LEMMA: In a $\triangle ABC$ let $P$ be a point and let $AP\cap \odot (BPC)=D$ and let $D^*$ be the reflection of $D$ in $BC$ then $AD^*$ passes through the antigonal conjugate of $P$ wrt $\triangle ABC$ . PROOF: Let $\mathcal{H}$ be the circumrectangular hyperbola of $\triangle ABC$ through $P$ and let $\ell$ be the isogonal conjugate of $\mathcal{H}$ wrt $\triangle ABC,$ let $Q$ be the antigonal conjugate of $P$ wrt $\triangle ABC$ Now fix $\mathcal{H}$ and vary $P$ so $P\mapsto Q$ is an involution on $\mathcal{H}$ with fixed points the isogonal conjugates of $\ell\cap \odot (ABC)$ now clearly as $P$ varies on $\mathcal{H}$ ,$D$ lies on a fixed circle $\Omega$ through $A,$reflection of $A$ in $BC$ (for instance see Small problem on fermat point and kiepert hyperbola(Post#4 ,LEMMA) so center of $\Omega$ lies on $BC$ so $D^*$ also lies on $\Omega$ now since $\Omega\cap BC$ are the images under $\sqrt{bc}$ inversion of $\ell\cap \odot (ABC)$ so if $X\in \Omega\cap BC$ then $AX$ is parallel to one of the asymptotes of $\mathcal{H}$ $(-\bigstar)$ now clearly the involution $AD\mapsto AD^*$ on $\Omega$ has fixed points $\Omega\cap BC$ so from $(\bigstar)$ we conclude that the involution $AP\mapsto AQ$ on $\mathcal{H}$ and the involution $AD\mapsto AD^*$ coincide so $AD^*$ passes through $Q$ the antigonal conjugate of $P$ wrt $\triangle ABC.$ ______________________________________________ Now let $I,I_A$ be the incenter$ ,A-$excenter of $\triangle ABC$ from LEMMA: we can immediately conclude that $I_BB'\cap I_CC'$ is the isogonal conjugate of antigonal conjugate of circumcenter of $\triangle I_AI_BI_C$ wrt $\triangle I_AI_BI_C$ that is nothing but $I^*$ the inverse of $I$ wrt $\triangle I_AI_BI_C$ now using LEMMA:2,LEMMA:3 in the above post we can conclude that $I^*$ lies on neuberg cubic of $\triangle ABC$ $,\implies P$ lies on neuberg cubic of $\triangle ABC$ so we are done $\qquad \blacksquare$

Here's a thoughtless bary bash (It just looks long ): Let $A'$ be the reflection of $A$ about $BC$, and let $I$ be the incenter of $\triangle ABC$. We start off with the following lemma- Lemma $P$ lies on $IA'$, i.e. $IA',I_BC',I_CB'$ are concurrent. PROOF: Note that we wish to show that $\triangle A'B'C'$ and $\triangle II_CI_B$ are perspective from a point. By Desargues' Theorem of Two Triangles, this is equivalent to showing that these two triangles are perspective from a line. We use barycentric coordinates wrt $\triangle ABC$. From here, we get that $$A'=(-a^2:X_C:X_B),B'=(X_C:-b^2:X_A),C'=(X_B:X_A:-c^2)$$where $X_A=b^2+c^2-a^2$, and $X_B$ and $X_C$ are defined similarly (Not using Conway's notation deliberately). Now, Let $II_B \cap A'C'=(a:t:c)$. Then, as this point lies on $A'C'$, we have \[\begin{vmatrix} a & t & c \\ -a^2 & X_C & X_B \\ X_B & X_A & -c^2 \end{vmatrix}=0 \]\[\ \Rightarrow a(c^2X_C+X_AX_B)+t(a^2c^2-X_B^2)+c(a^2X_A+X_BX_C)=0 \]\[\ \Rightarrow ac(aX_A+cX_C)+X_B(aX_A+cX_C)+t(ac-X_B)(ac+X_B)=0 \]\[\ \Rightarrow (X_B+ac)(aX_A+cX_C)=t(X_B+ac)(X_B-ac) \Rightarrow t=\frac{aX_A+cX_C}{X_B-ac} \]So we have $$II_B \cap A'C'=(a(X_B-ac):aX_A+cX_C:c(X_B-ac))$$Similarly, we get $$II_C \cap A'B'=(a(X_C-ab):b(X_C-ab):aX_A+bX_B)$$And, taking $I_BI_C \cap B'C'=(k:b:-c)$, and by doing a similar analysis (which was quite symmetric, and so I didn't really do it again), we get that $$I_BI_C \cap B'C'=(bX_B-cX_C:b(X_A+bc):-c(X_A+bc))$$Now, we just need to show that these three points are collinear, or equivalently that the following determinant is zero. \[\begin{vmatrix} a(X_B-ac) & aX_A+cX_C & c(X_B-ac) \\ a(X_C-ab) & b(X_C-ab) & aX_A+bX_B \\ bX_B-cX_C & b(X_A+bc) & -c(X_A+bc) \end{vmatrix}=0 \]i.e. \[\ -a(X_B-ac) \times b(X_A+bc)(c(X_C-ab)+(aX_A+bX_B))+(aX_A+cX_C)(ac(X_C-ab)(X_A+bc)+(aX_A+bX_B)(bX_B-cX_C))+c(X_B-ac) \times b(X_C-ab)(a(X_A+bc)-(bX_B-cX_C))=0 \]i.e. \[\ (bX_B-cX_C)[(aX_A+cX_C)(aX_A+bX_B)-bc(X_B-ac)(X_C-ab)]=a(X_A+bc)[b(X_B-ac)(aX_A+bX_B)-c(X_C-ab)(aX_A+cX_C)] \]Now, We make use of the following transformations ($s$ is the semiperimeter). \[\ aX_A+cX_C=a(b^2+c^2-a^2)+c(a^2+b^2-c^2)=(a+c)(b+c-a)(a+b-c)=4(a+c)(s-a)(s-c) \]\[\ bX_B-cX_C=b(a^2+c^2-b^2)-c(a^2+b^2-c^2)=(c-b)(a+b+c)b+c-a)=4s(c-b)(s-a) \]\[\ X_B-ac=a^2+c^2-b^2-ac=ac-(b+c-a)(a+b-c)=ac-4(s-a)(s-c) \]Then we have that \begin{align*} LHS &= 4s(c-b)(s-a)[a^2X_A^2+aX_A(bX_B+cX_C)+abc(bX_B+cX_C)-a^2b^2c^2] \\ &= 4as(c-b)(s-a)[a(X_A+bc)(X_A-bc)+(X_A+bc)(bX_B+cX_C)] \\ &= 4as(s-a)(c-b)(X_A+bc)[a\{bc-4(s-b)(s-c)\}+4(b+c)(s-b)(s-c)] \\ &= 4as(s-a)(c-b)(X_A+bc)[abc+4(s-b)(s-c)(b+c-a)] \\ &= 4as(s-a)(c-b)(X_A+bc)[abc+8(s-a)(s-b)(s-c) \\ \\ RHS &= a(X_A+bc)[4b(a+b)(s-a)(s-b) \times \{ac-4(s-a)(s-c)\}-4c(a+c)(s-a)(s-c)\{ab-4(s-a)(s-b)\}] \\ &= 4a(s-a)(X_A+bc)[abc\{(a+b)(s-b)-(a+c)(s-c)\}-4(s-a)(s-b)(s-c)\{b(a+b)-c(a+c)\}] \\ &= 4a(s-a)(X_A+bc)[abc(c-b)(a+b+c-s)-4(s-a)(s-b)(s-c) \times (c-b)(a+b+c)] \\ &= 4a(s-a)(X_A+bc) \times s(c-b)[abc+8(s-a)(s-b)(s-c)] \\ \end{align*}Thus, $LHS=RHS$, proving our Lemma. $\Box$ Return to the problem at hand. As $BB'PP_B$ is an isosceles trapezoid, we have that $AC,B'P,BP_B$ concur. Similarly, $AB,C'P,CP_C$ and $AP_A,A'P,BC$ also concur. Taking $A'I \cap BC=X,I_CB' \cap CA=Y,I_BC' \cap AB=Z$, we wish to show that $AX,BY,CZ$ are concurrent. Now, Let $Y=(1:0:m)$. Then, as $Y$ lies on $I_CB'$, we have \[\begin{vmatrix} 1 & 0 & m \\ X_C & -b^2 & X_A \\ a & b & -c \end{vmatrix}=0 \]\[\ \Rightarrow (b^2c-bX_A)+m(bX_C+ab^2)=0 \Rightarrow m=\frac{X_A-bc}{X_C+ab} \] This gives $Y=(X_C+ab:0:X_A-bc)$. By symmetry, we get that $Z=(X_B+ac:X_A-bc:0)$. Again, let $X=(0:p:1)$. Then, by a similar analysis, one can easily find out that $X=(0:X_C+ab:X_B+ac)$. Thus, $$\frac{BX}{CX} \times \frac{CY}{AY} \times \frac{AZ}{BZ}=\frac{X_B+ac}{X_C+ab} \times \frac{X_C+ab}{X_A-bc} \times \frac{X_A-bc}{X_B+ac}=1$$By the Converse of Ceva's Theorem, we get that $AX,BY,CZ$ concur. Hence, done. $\blacksquare$

Here is an approach that I find easier to be understood. Let $A'$ be the reflection of $A$ over $BC$, and $I_A$ the excenter opposite of $A$. Denote the angles of the triangle as $2a,2b,2c$. First of al,triangles $\triangle II_BI_C$ and $\triangle A'B'C'$ are perspective. Let $II_B\cap A'C'=B_1$ $II_C\cap A'B'=C_1$ $B'C'\cap I_CI_B=A_1$. Notice that it is enough to prove that those points are collinear. By the symmetry,it is obvious that $BB_1,CC_1$ are the angle bisectors of $C'BA'$ and $A'CB'$. So we have $\frac{A'B_1}{B_1C'}=\frac {A'B}{C'B}$ and $ \frac{B'C_1}{A'C_1}=\frac {B'C}{A'C}$. Also observe that $I_BI_C$ is the external bisector of the angle $\angle C'AB'$ so $\frac{A_1C'}{A_1B'}=\frac {AC'}{AB'}$. Multiplying those, by Menelaos, we get the collinearity and also by Desargues Theorem that $A',I,P$ are collinear. Now notice that by symmetry $\angle ABP_B=\angle AB'P$, $\angle CBP_B=\angle CB'P$. By Sine Law in $I_CCB',I_CAB',I_CAC$ we get: $\frac{sin(P_BBC)}{sin(P_BBA)}=\frac{sin(AB'P)}{sin(CB'P)}=\frac{sin(3c)sin(90+a)}{sin(c)sin(90+3a)}$. Analogeously, we get $\frac{sin(P_CCB)}{sin(P_BCA)}=\frac{sin(90+3a)sin(b)}{sin(90+a)sin(3b)}$. For $P_A$, by symmetry and the Sine Law in $IA'C,IA'B$ we have: $\frac{sin(P_AAC)}{sin(P_AAB)}=\frac{sin(CA'P)}{sin(CA'B)}=\frac{sin(3b)sin(c)}{sin(3c)sin(b)}$. Multiplying the last 3 relations and by Trig Ceva, we get $AP_A,BP_B,CP_C$ concurrent, as desired.

Can anyone simplify what Cayley-Bacharach Theorm means? on the net it says that if two cubics have 8 points common then they have a nine'th point common too but I dont see how @TelvCohl used this in her generalization

tastymath75025 wrote: Note: can someone please elaborate on TelvCohl's proof that $P^* \in I_CB', I_BC', IA'$? The proof I gave was the fastest one I could come up with Denote $ \mathbf{I} $ as the inversion with center $ I $ that exchanges $ A, I_A. $ Since $ A' $ is the reflection of $ A $ in $ BC, $ so $ \mathbf{I}(A') $ is the image of $ \mathbf{I}(A)=I_A $ under the inversion WRT $ \mathbf{I}(\overline{BC}) = \odot (II_BI_C). $

Copy USAMO 2016 P3 Let $I, I_A$ be the incenter and the $A$-excenter of $\triangle ABC$. Let $A'$ be the reflection of $A$ across $BC$. Step 1: $P,I,A'$ are colinear. Let $A$-excircle of $\triangle ABC$ touches $BC, CA, AB$ at $D,E,F$ and let $\triangle D_1E_1F_1$ be the image of $\triangle DEF$ under the homothety $\mathcal{H}(I_A, 4)$. By the external version of "midpoint of altitude lemma", we get that $I_CE$ pass through midpoint of $B$-altitude. Hence $I_C, E_1, B'$ are colinear. Similarly, $I_B, F_1, C'$ are colinear and $I, D_1, A'$ are colinear. Hence $IA', I_CB', I_BC'$ are concurrent at the homothety center of $\triangle II_CI_B$ and $\triangle D_1E_1F_1$. Remark: One can also show easily that $P\in OI_A$, resembling USAMO 2016 P3. Step 2 Finishing. Let $I_C', I_B', I'$ be the reflection of $I_C, I_B, I$ across $AC, AB, BC$ respectively. As $P_A\in AI'$ and similar relations, it suffices to show that $AI', BI_C', CI_B'$ are concurrent. Let $\triangle D_2E_2F_2$ be the image of $\triangle DEF$ under the homothety $\mathcal{H}(I_A, -2)$. By homothety, the line connecting $B$ and the tangency point between $C$-excircle and $AC$ pass through the antipode of $E$ w.r.t. $A$-excircle. Therefore $B, I_C', E_2$ are colinear. Similarly $\{A, I', D_2\}$ and $\{C, I_B', F_2\}$ are colinear. Hence the problem boils down to showing that $AD_2, BE_2, CF_2$ are concurrent. But observe that $\triangle ABC$ and $\triangle D_2E_2F_2$ are orthologic with orthology centers coincide at $I_A$, hence by Sondat's theorem, these triangle must be perspective so we are done. TinaSprout wrote: tastymath75025 wrote: Note: can someone please elaborate on TelvCohl's proof that $P^* \in I_CB', I_BC', IA'$? The proof I gave was the fastest one I could come up with Denote $ \mathbf{I} $ as the inversion with center $ I $ that exchanges $ A, I_A. $ Since $ A' $ is the reflection of $ A $ in $ BC, $ so $ \mathbf{I}(A') $ is the image of $ \mathbf{I}(A)=I_A $ under the inversion WRT $ \mathbf{I}(\overline{BC}) = \odot (II_BI_C). $ Can you explain in more detail? I don't get what's going on here.

MarkBcc168 wrote: TinaSprout wrote: tastymath75025 wrote: Note: can someone please elaborate on TelvCohl's proof that $P^* \in I_CB', I_BC', IA'$? The proof I gave was the fastest one I could come up with Denote $ \mathbf{I} $ as the inversion with center $ I $ that exchanges $ A, I_A. $ Since $ A' $ is the reflection of $ A $ in $ BC, $ so $ \mathbf{I}(A') $ is the image of $ \mathbf{I}(A)=I_A $ under the inversion WRT $ \mathbf{I}(\overline{BC}) = \odot (II_BI_C). $ Can you explain in more detail? I don't get what's going on here. In general, if $ P, Q $ are inverse WRT a circle $ \mathcal{C}, $ then $ \phi (P), \phi (Q) $ are inverse WRT the circle $ \phi (\mathcal{C}), $ where $ \phi $ is an inversion. In the original problem we consider $ A' $ as the image of $ A $ under the inversion WRT circle $ \overline{BC}. $

Construct the Neuberg cubic $\mathcal C,$ the locus of points whose reflection triangle is perspective with $\triangle ABC,$ which passes through $A,B,C,I,I_A,I_B,I_C,A',B',C',H.$ It suffices to show $P$ lies on this cubic. This follows by Cayley-Bacharach, since the two cubics $\overline{BHB'}\cup\overline{CII_C}\cup\overline{I_BC'P}$ and $\overline{BII_B}\cup\overline{CHC'}\cup\overline{B'I_CP}$ meet at the nine points $B,C,H,B',C',I,I_B,I_C,P,$ the first eight of which lie on $\mathcal C$ and thus the ninth does as well.